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Question: Solve: \(\sec \theta -1=\left( \sqrt{2}-1 \right)\tan \theta \)....

Solve: sec⁡θ−1=(2−1)tan⁡θ\sec \theta -1=\left( \sqrt{2}-1 \right)\tan \theta .

Explanation

Solution

We use the given equation and factorise it into multiplication of two trigonometric functions by converting them into terms of sine and cosine function. We solve that to find two equational parts. We find their general solution and take the combined solution.

Complete step by step answer:
We apply the theorems of trigonometry cos⁡θ.sec⁡θ=1,tan⁡θ=sin⁡θcos⁡θ\cos \theta .\sec \theta =1,\tan \theta =\dfrac{\sin \theta }{\cos \theta }. The equation becomes
sec⁡θ−1=(2−1)tan⁡θ ⇒cos⁡θ(sec⁡θ−1)=(2−1)tan⁡θ.cos⁡θ ⇒1−cos⁡θ=(2−1)sin⁡θ \begin{aligned} & \sec \theta -1=\left( \sqrt{2}-1 \right)\tan \theta \\\ & \Rightarrow \cos \theta \left( \sec \theta -1 \right)=\left( \sqrt{2}-1 \right)\tan \theta .\cos \theta \\\ & \Rightarrow 1-\cos \theta =\left( \sqrt{2}-1 \right)\sin \theta \\\ \end{aligned}
Now we apply formula of submultiple angles which tells us
1−cos⁡θ=2sin⁡2θ2,sin⁡θ=2sin⁡θ2cos⁡θ21-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2},\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2} .
Now the equation changes to
1−cos⁡θ=(2−1)sin⁡θ ⇒2sin⁡2θ2=(2−1)(2sin⁡θ2cos⁡θ2) ⇒2sin⁡θ2(sin⁡θ2−(2−1)cos⁡θ2)=0 \begin{aligned} & 1-\cos \theta =\left( \sqrt{2}-1 \right)\sin \theta \\\ & \Rightarrow 2{{\sin }^{2}}\dfrac{\theta }{2}=\left( \sqrt{2}-1 \right)\left( 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2} \right) \\\ & \Rightarrow 2\sin \dfrac{\theta }{2}\left( \sin \dfrac{\theta }{2}-\left( \sqrt{2}-1 \right)\cos \dfrac{\theta }{2} \right)=0 \\\ \end{aligned}
We converted the given equation into its factorisation form. We break it into two parts as
2sin⁡θ2=0,(sin⁡θ2−(2−1)cos⁡θ2)=02\sin \dfrac{\theta }{2}=0,\left( \sin \dfrac{\theta }{2}-\left( \sqrt{2}-1 \right)\cos \dfrac{\theta }{2} \right)=0.
The first equation gives one part of the general solution for θ\theta .
2sin⁡θ2=0⇒sin⁡θ2=0 ⇒θ2=nπ ⇒θ=2nπ,[n∈Z]..........(i) \begin{aligned} & 2\sin \dfrac{\theta }{2}=0\Rightarrow \sin \dfrac{\theta }{2}=0 \\\ & \Rightarrow \dfrac{\theta }{2}=n\pi \\\ & \Rightarrow \theta =2n\pi ,\left[ n\in \mathbb{Z} \right]..........(i) \\\ \end{aligned}
Now we work with the second part (sin⁡θ2−(2−1)cos⁡θ2)=0\left( \sin \dfrac{\theta }{2}-\left( \sqrt{2}-1 \right)\cos \dfrac{\theta }{2} \right)=0. Solving this we get
(sin⁡θ2−(2−1)cos⁡θ2)=0 ⇒sin⁡θ2=(2−1)cos⁡θ2 ⇒tan⁡θ2=(2−1) \begin{aligned} & \left( \sin \dfrac{\theta }{2}-\left( \sqrt{2}-1 \right)\cos \dfrac{\theta }{2} \right)=0 \\\ & \Rightarrow \sin \dfrac{\theta }{2}=\left( \sqrt{2}-1 \right)\cos \dfrac{\theta }{2} \\\ & \Rightarrow \tan \dfrac{\theta }{2}=\left( \sqrt{2}-1 \right) \\\ \end{aligned}
We got the value of tan⁡θ2\tan \dfrac{\theta }{2}. Using the submultiple angle formula tan⁡θ=2tan⁡θ21−tan⁡2θ2\tan \theta =\dfrac{2\tan \dfrac{\theta }{2}}{1-{{\tan }^{2}}\dfrac{\theta }{2}}, we get the value of tan⁡θ\tan \theta .
So, tan⁡θ=2tan⁡θ21−tan⁡2θ2=2(2−1)1−(2−1)2=2(2−1)1−2−1+22=2(2−1)2(2−1)=1\tan \theta =\dfrac{2\tan \dfrac{\theta }{2}}{1-{{\tan }^{2}}\dfrac{\theta }{2}}=\dfrac{2\left( \sqrt{2}-1 \right)}{1-{{\left( \sqrt{2}-1 \right)}^{2}}}=\dfrac{2\left( \sqrt{2}-1 \right)}{1-2-1+2\sqrt{2}}=\dfrac{2\left( \sqrt{2}-1 \right)}{2\left( \sqrt{2}-1 \right)}=1
Now we use the formula of general solution to find another set of value for θ\theta .

& \tan \theta =1=\tan \left( \dfrac{\pi }{4} \right) \\\ & \Rightarrow \theta =n\pi +\dfrac{\pi }{4},\left[ n\in \mathbb{Z} \right]...................(ii) \\\ \end{aligned}$$ **From solution (i) and (ii) we get the final solution of the equation $\sec \theta -1=\left( \sqrt{2}-1 \right)\tan \theta $ as $$\theta =\left( 2n\pi \right)\cup \left( n\pi +\dfrac{\pi }{4} \right),\left[ n\in \mathbb{Z} \right]$$.** **Note:** We can’t eliminate the part of $2\sin \dfrac{\theta }{2}$ from the factorisation $2\sin \dfrac{\theta }{2}\left( \sin \dfrac{\theta }{2}-\left( \sqrt{2}-1 \right)\cos \dfrac{\theta }{2} \right)=0$ as that includes one solution of the main equation. $2\sin \dfrac{\theta }{2}=0$ also satisfies the equation.